MICROELECTRONICS and NANOTECHNOLOGY (MN)

Question 2: Junction Devices

August 2013

## Questions

All questions are in this link

# Solutions of all questions

1) Two effects: i) Kirk effect or base pushout ii) Current crowding

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2) $Eg(InP) > Eg(InGaAs)$

 So; InP is in emitter and collector.


• $\beta$ increases as:

\begin{align*} \beta &=\frac{D_n}{D_p}\cdot\frac{W_E}{W_B}\cdot\frac{n_{iB}^2}{n_{iE}^2}\cdot\frac{B_E}{N_B}\\ &\approx\frac{n_{iB}^2}{n_{iE}^2}\cdot\frac{B_E}{N_B}\cdot\frac{V_{th}}{v_s} \end{align*} $\frac{n_{iB}^2}{n_{iE}^2} = e^{\triangle Eg/kT}$

So huge improvement

• $V_{cb,br}$ increases as $V_{cb,br}\approx\frac{3}{2}$ Eg collector

• $f_{max}$ increases

$f_{max}^{-1}\approx f_T^{-1} = \bigg(\frac{w_B^2}{2D_n}+\frac{w_{BC}}{2v_{sat}}\bigg)+\frac{kT}{qSe}[C_{jBC}+C_{jBE}]$ here; $w_B\downarrow I_C\uparrow C_J\downarrow$ all leads to $f_T\uparrow$

• high current effect reduces

• $V_cesat$ (??)
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3)a)\begin{align*} V_a&=0\\ \therefore E&=0 \end{align*}


$\therefore n_p(x,t) = \frac{N}{\sqrt{4\pi D_nt}}exp\bigg(\frac{-x^2}{4D_nt}-\frac{t}{\tau_n}\bigg)+n_{p0}$

 At $x=0$


2 unknowns $\to D_n,\tau_n$ selecting 2 specific times and using the excess carrier can find $D_n$ and $\tau_n$.

At $x=X$; initially no excess carrier.

b) For applied field; decay will be much quicker.

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## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva